Asymptotically Stable Solutions of a Generalized Fractional Quadratic Functional-Integral Equation of Erdélyi-Kober Type

Abstract

We study a generalized fractional quadratic functional-integral equation of Erdélyi-Kober type in the Banach space . We show that this equation has at least one asymptotically stable solution.

1. Introduction

Quadratic integral equations with nonsingular kernels have received a lot of attention because of their useful applications in describing numerous events and problems of the real world. For example, quadratic integral equations are often applicable in kinetic theory of gases, in the theory of neutron transport, and in the traffic theory; see [1–8]. The existence of solutions for several classes of nonlinear quadratic integral equations with nonsingular kernels has been studied by several authors, for example, Argyros [9], Banaś et al. [10–12], Benchohra and Darwish [13, 14], Caballero et al. [15–17], Darwish et al. [18, 19], Leggett [20], and Stuart [21]. There is a great interest in studying singular quadratic integral equations by many authors, after the appearance of Darwish’s paper [22], for example, Banaś and O’Regan [23], Banaś and Rzepka [24, 25], Darwish [26, 27], Darwish and Sadarangani [28], Darwish and Ntouyas [29], Darwish et al. [30], and Wang et al. [31, 32].

In this paper, we will study the quadratic functional-integral equation of fractional order where and .

If and , we obtain a quadratic Urysohn-Volterra integral equation of fractional order studied by Banas’ and O’Regan in [23] while in the case where , , and , we get a fractional quadratic integral equation of Hammerstein-Volterra type studied by Darwish in [22]. Moreover, in the case where , we obtain the quadratic functional-integral equation of fractional order studied by Darwish and Sadarangani in [28].

The aim of this paper is to prove the existence of solutions of (1) in the space of real functions, defined, continuous, and bounded on an unbounded interval. Moreover, we will obtain some asymptotic characterization of solutions of (1). Our proof depends on suitable combination of the technique of measures of noncompactness and the Schauder fixed point principle.

2. Notation and Auxiliary Facts

This section is devoted to collecting some definitions and results which will be needed further on. First, we recall from [33–35] that the Erdélyi-Kober fractional integral of a continuous function is defined as When , we obtain Riemann-Liouville fractional integral; that is,

Now, let be an infinite dimensional Banach space with zero element . Let denote the closed ball centered at with radius . The symbol stands for the ball .

If is a subset of , then and denote the closure and convex closure of , respectively. Moreover, we denote by the family of all nonempty and bounded subsets of and by its subfamily consisting of all relatively compact subsets.

Next we give the definition of the concept of a measure of noncompactness [36].

Definition 1. A mapping is said to be a measure of noncompactness in if it satisfies the following conditions. (1)The family is nonempty and .(2).(3).(4) for .(5)If , , for and , then .

The family described above is called the kernel of the measure of noncompactness . Let us observe that the intersection set from belongs to . In fact, since for every, then we have that .

In what follows we will work in the Banach space consisting of all real functions defined, bounded, and continuous on . This space is equipped with the standard norm

Next, we give the construction of the measure of noncompactness in which will be used as main tool of the proof of our main result; see [37, 38] and references therein.

Let us fix a nonempty and bounded subset of and numbers and . For arbitrary function let us denote by the modulus of continuity of the function on the interval ; that is, Further, let us put Moreover, for a fixed number let us define Let us mention that the kernel consists of all nonempty and bounded sets such that functions belonging to are locally equicontinuous on . On the other hand, the kernel is the family containing all nonempty and bounded sets in the space such that the thickness of the bundle formed by the graphs of functions belonging to tends to zero at infinity.

Finally, with the help of the above quantities we can define a measure of noncompactness as The function is a measure of noncompactness in the space [36, 37].

In the end of this section, we recall the definition of the asymptotic stability solutions which will be used in the proof of our main result. To this end we assume that is a nonempty subset of the space . Let be a given operator. We consider the following operator equation:

Definition 2. One says that solutions of (9) are asymptotically stable if there exists a ball such that and such that for each there exists such that for arbitrary solutions ,   of this equation belonging to the inequality is satisfied for any .

3. The Existence and Asymptotic Stability of Solutions

In this section we will study (1) assuming that the following hypotheses are satisfied. is a continuous and bounded function on .   is continuous and the function is bounded on with . Moreover, there exists a continuous function such that  for all and for any .  is continuous and there exists a continuous function such that  for all and for any .  is a continuous function. Moreover, there exist a function being continuous on and a function being continuous and nondecreasing on with such that  for all such that and for all . For further purpose let us define the function by . The functions defined by , , , and are bounded on and the functions and vanish at infinity; that is, . There exists a positive solution of the inequality  and , where , , , and .

Now, we are in a position to state and prove our main result.

Theorem 3. Let the hypotheses be satisfied. Then (1) has at least one solution and all solutions of this equation belonging to the ball are asymptotically stable.

Proof. Denote by the operator associated with the right-hand side of (1). Then, (1) takes the form where Here, and are the superposition operators, generated by the functions and involved in (1), defined by respectively, where is an arbitrary function defined on (see [39]).
Solving (1) is equivalent to finding a fixed point of the operator defined on the space .
For convenience, we divide the proof into several steps.
Step  1 (  is continuous on ). To prove the continuity of the function on it suffices to show that if , then is continuous function on , thanks to , , and . For this purpose, take an arbitrary and fix and . Assume that are such that . Without loss of generality we can assume that . Then we get Let us denote then we obtain Thus where
In view of the uniform continuity of the function on we have that as . From the above inequality we infer that the function is continuous on the interval for any . This yields the continuity of on and, consequently, the function is continuous on .
Step  2 ( is bounded on ). In view of our hypotheses for arbitrary and for a fixed we have Hence, is bounded on , thanks to hypothesis .
Step  3  ( maps the ball into itself). Steps and allow us to conclude that the operator transforms into itself. Moreover, from the last estimate we have From the last estimate with hypothesis we deduce that there exists such that the operator maps into itself.
Step  4 (an estimate of with respect to the quantity ). Let us take a nonempty set . Then, for arbitrary and for a fixed , we obtain Hence, we can easily deduce the following inequality: Now, taking into account hypothesis we obtain where . Obviously, in view of hypothesis we have that .
Step  5 (an estimate of with respect to the modulus of continuity ). Take arbitrary numbers and . Choose a function and take such that . Without loss of generality we can assume that . Then, taking into account our hypotheses and (21), we have In the last estimates, we have denoted by Hence, Since the function is uniformly continuous on the set , the function is uniformly continuous on the set and the function is uniformly continuous on the set , where we have It is easy to see that because is bounded on , is bounded on , and .
Therefore, from the last estimate we derive the following one:
Hence we have
Step  6 ( is contraction with respect to the measure of noncompactness ). From (27) and (34) and the definition of the measure of noncompactness given by formula (8), we obtain
Step  7. We construct a nonempty, bounded, closed, and convex set on which we will apply a fixed point theorem. In the sequel let us put , , and so on. In this way we have constructed a decreasing sequence of nonempty, bounded, closed, and convex subsets of such that for . Hence, in view of (35) we obtain This implies that . Hence, taking into account Definition 1 we infer that the set is nonempty, bounded, closed, and convex subset of . Moreover, . Also, the operator maps into itself.
Step  8 ( is continuous on the set ). Let us fix a number and take arbitrary functions such that . Using the fact that and keeping in mind the structure of sets belonging to we can find a number such that for each and we have that . Since maps into itself, we have that . Thus, for we get On the other hand, let us assume . Then we obtain Now, taking into account (37) and (38) and hypothesis we conclude that the operator is continuous on the set .
Step  9 (application of Schauder fixed point principle). Linking all above-obtained facts about the set and the operator and using the classical Schauder fixed point principle we deduce that the operator has at least one fixed point in the set . Obviously the function is a solution of the quadratic integral equation (1). Moreover, since we have that all solutions of (1) belonging to are asymptotically stable in the sense of Definition 2. This completes the proof.

4. Example

In this section, we present an example as an application of Theorem 3.

Consider the following integral equation of fractional order: Equation (39) is a special case of (1), where , , is a positive constant, and

It is easy to check that the assumptions of Theorem 3 are satisfied. In fact we have that the function is continuous and bounded on and .

The function satisfies assumption with and , being .

Moreover, the function satisfies assumption with .

The function satisfies assumption with , , , and .

Next, we are going to check that assumption is satisfied. The functions , , , and appearing in assumption take the form It is easy to see that .

Moreover we have , , and .

Therefore the inequality in assumption has the form We can easily check that the number is a solution of the inequality (43) for . Now, by Theorem 3, we infer that our equation has a solution in and all solutions of (39) which belongs to are asymptotically stable in the sense of the Definition 2.

Conflict of Interests

The authors declare that there is no conflict of interests in the submitted paper.